1. ## Log Return Question

New to the forum. Math has been a lifetime of frustration for me so apologies if I don't articulate this question using the appropriate terminology. I know that if you take the ln(x) - ln(x-1) this is the log return. I think 1 + this log return would be considered comparable to a simple return of x / x-1. The question I came here for is what does ln(x) / ln(x-1) tell you??

Thanks for your time and help.

2. ## Re: Log Return Question

$\displaystyle \ln(x) - \ln(x-1)$

$\displaystyle =\ln(x) - \ln(x-1)$

$\displaystyle =\ln\left(\frac{x}{x-1}\right)$

$\displaystyle 1 + \log\left(\ln\left(\frac{x}{x - 1}\right) \right) \ne \frac{x}{x-1}$

You can check what I just stated here. Wolfram|Alpha: Computational Intelligence

And $\displaystyle \frac{\ln(x)}{\ln(x-1)} = \frac{\text{log of x base e}}{\text{log of (x - 1) base e}}$

Also $\displaystyle \frac{\ln(x)}{\ln(x-1)} = \log_{(x-1)} {x}$

http://www.wolframalpha.com/input/?i...+log_(x-1)+(x)

I hope it helps.